An airplane is cruising at 340 km/h at an altitude of 4000 m. If the plane passes directly over an observer on the ground, how fast is the distance from the plane to the observer changing when it is 3000 m away from the point directly above the observer?
if the distance from the plane to the observer is y, and the distance the plane has flown is x, all in km,
y^2 = 4^2 + x^2
2y dy/dt = 2x dx/dt
when x = 3, y=5
10 dy/dt = 2(3)(340)
dy/dt = 204 km/hr
To find the rate at which the distance between the plane and the observer is changing, we can use the concept of related rates.
Let's denote the horizontal distance between the point directly above the observer and the plane as x, and the vertical distance between them as y.
We know that the plane is moving horizontally at a constant speed of 340 km/h, which can be converted to meters per hour:
340 km/h = 340,000 m/h.
Since the plane is moving horizontally, the rate of change of x with respect to time is constant at 340,000 m/h.
At a particular moment, we want to find the rate at which y is changing with respect to time (dy/dt). To do this, we can use the Pythagorean theorem:
x^2 + y^2 = d^2
where d is the distance between the plane and the observer. In this case, d is equal to 3000 m since the plane is 3000 m away from the point directly above the observer.
Differentiating both sides of the equation with respect to time gives:
2x(dx/dt) + 2y(dy/dt) = 2d(dd/dt)
Since dx/dt is the horizontal speed of the plane, which is constant at 340,000 m/h, and d is constant at 3000 m, the equation simplifies to:
2x(340,000) + 2y(dy/dt) = 0
Now, we need to find the value of y when x is 3000 m away from the point directly above the observer. Using the Pythagorean theorem, we have:
3000^2 + y^2 = 4000^2
Solving for y, we get:
y^2 = 4000^2 - 3000^2
y^2 = 16,000,000 - 9,000,000
y^2 = 7,000,000
y = √7,000,000
y ≈ 2645.75
Now, we substitute the values into the equation:
2(3000)(340,000) + 2(2645.75)(dy/dt) = 0
Rearranging the equation and solving for dy/dt, we get:
2(3000)(340,000) = -2(2645.75)(dy/dt)
dy/dt = -2(3000)(340,000) / (2)(2645.75)
dy/dt ≈ -2,040,000 m/h
Therefore, when the plane is 3000 m away from the point directly above the observer, the distance from the plane to the observer is changing at a rate of approximately -2,040,000 m/h. The negative sign indicates that the distance is decreasing.